Problem: $\dfrac{ -q + 7r }{ -5 } = \dfrac{ 9q - 3s }{ 4 }$ Solve for $q$.
Solution: Multiply both sides by the left denominator. $\dfrac{ -q + 7r }{ -{5} } = \dfrac{ 9q - 3s }{ 4 }$ $-{5} \cdot \dfrac{ -q + 7r }{ -{5} } = -{5} \cdot \dfrac{ 9q - 3s }{ 4 }$ $-q + 7r = -{5} \cdot \dfrac { 9q - 3s }{ 4 }$ Multiply both sides by the right denominator. $-q + 7r = -5 \cdot \dfrac{ 9q - 3s }{ {4} }$ ${4} \cdot \left( -q + 7r \right) = {4} \cdot -5 \cdot \dfrac{ 9q - 3s }{ {4} }$ ${4} \cdot \left( -q + 7r \right) = -5 \cdot \left( 9q - 3s \right)$ Distribute both sides ${4} \cdot \left( -q + 7r \right) = -{5} \cdot \left( 9q - 3s \right)$ $-{4}q + {28}r = -{45}q + {15}s$ Combine $q$ terms on the left. $-{4q} + 28r = -{45q} + 15s$ ${41q} + 28r = 15s$ Move the $r$ term to the right. $41q + {28r} = 15s$ $41q = 15s - {28r}$ Isolate $q$ by dividing both sides by its coefficient. ${41}q = 15s - 28r$ $q = \dfrac{ 15s - 28r }{ {41} }$